Recover Rotated Sorted Array

Question

• lintcode: (39) Recover Rotated Sorted Array

Given a rotated sorted array, recover it to sorted array in-place.

Example
[4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5]

Challenge
In-place, O(1) extra space and O(n) time.

Clarification
What is rotated array:

    - For example, the orginal array is [1,2,3,4], The rotated array of it can be [1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]

首先可以想到逐步移位，但是这种方法显然太浪费时间，不可取。下面介绍利器『三步翻转法』，以[4, 5, 1, 2, 3]为例。

1. 首先找到分割点5和1
2. 翻转前半部分4, 5为5, 4，后半部分1, 2, 3翻转为3, 2, 1。整个数组目前变为[5, 4, 3, 2, 1]
3. 最后整体翻转即可得[1, 2, 3, 4, 5]

由以上3个步骤可知其核心为『翻转』的in-place实现。使用两个指针，一个指头，一个指尾，使用for循环移位交换即可。

• reference
• reference
• visualization

class Solution:
    """
    @param nums: The rotated sorted array
    @return: nothing
    """
    def recoverRotatedSortedArray(self, nums):
        # write your code here
        if nums==None:
            return
        numslen = len(nums)
        for index in range(numslen-1):
            if(nums[index]>nums[index+1]):
                self.reverse(nums,0,index)
                self.reverse(nums,index+1,numslen-1)
                self.reverse(nums,0,numslen-1)

    def reverse(self,nums,start,end):
        while start<end:
            nums[start],nums[end]=nums[end],nums[start]
            start+=1
            end-=1
    '''        
    def swap(self,nums,start,end):
        tmp = nums[start]
        nums[start] = nums[end]
        nums[end] = tmp
    '''

sol = Solution()
nums = [4, 5, 1, 2, 3] 
sol.recoverRotatedSortedArray(nums)